Find the domain of the following functions: (a) 𝑓(𝑥) = 1
√𝑥 + |𝑥|
(b) 𝑓(𝑥) = 1 . √[𝑥]2 − [𝑥] − 6

Let's analyze each function to find its domain.

### (a) \( f(x) = \frac{1}{\sqrt{x} + |x|} \)

For this function, the denominator \( \sqrt{x} + |x| \) must be defined and non-zero. Here are the considerations:

1. **Domain of the square root function** \( \sqrt{x} \): For the square root function to be real, \( x \geq 0 \).
2. **Absolute value function** \( |x| \): Defined for all \( x \), so no restrictions here.
3. **Denominator non-zero**: We need \( \sqrt{x} + |x| \neq 0 \).

Let's analyze \( \sqrt{x} + |x| \):
- For \( x \geq 0 \), \( |x| = x \), so the expression becomes \( \sqrt{x} + x \). This is always greater than zero for \( x \geq 0 \), meaning the denominator is non-zero for \( x \geq 0 \).
- For \( x < 0 \), \( |x| = -x \), but \( \sqrt{x} \) is undefined for negative \( x \).

Thus, the domain for this function is:
\[
\boxed{x > 0}
\]
(Since \( x = 0 \) makes the denominator zero, it is excluded.)

---

### (b) \( f(x) = \frac{1}{\sqrt{x^2 - x - 6}} \)

For this function, we must ensure the denominator \( \sqrt{x^2 - x - 6} \) is defined and non-zero. Therefore:

1. **Square root domain**: The expression inside the square root, \( x^2 - x - 6 \), must be non-negative, i.e., \( x^2 - x - 6 \geq 0 \).
2. **Denominator non-zero**: \( \sqrt{x^2 - x - 6} \neq 0 \).

Let's solve the inequality \( x^2 - x - 6 \geq 0 \):

- Factor the quadratic: \( x^2 - x - 6 = (x - 3)(x + 2) \).
- Solve \( (x - 3)(x + 2) \geq 0 \).

To solve this inequality, determine the sign of the expression by testing intervals based on the roots \( x = 3 \) and \( x = -2 \):
- For \( x < -2 \), both factors are negative, so the product is positive.
- For \( -2 < x < 3 \), one factor is positive, the other is negative, so the product is negative.
- For \( x > 3 \), both factors are positive, so the product is positive.

Thus, \( x^2 - x - 6 \geq 0 \) for \( x \in (-\infty, -2] \cup [3, \infty) \).

Now, exclude points where \( x^2 - x - 6 = 0 \) (i.e., \( x = -2 \) and \( x = 3 \)), as they make the denominator zero.

Thus, the domain is:
\[
\boxed{(-\infty, -2) \cup (3, \infty)}
\]

---

### Summary of Domains:
- (a) \( f(x) = \frac{1}{\sqrt{x} + |x|} \): Domain is \( x > 0 \).
- (b) \( f(x) = \frac{1}{\sqrt{x^2 - x - 6}} \): Domain is \( (-\infty, -2) \cup (3, \infty) \).

Would you like a more detailed explanation of any part?

### Related Questions:
1. What is the domain of \( f(x) = \frac{1}{x + |x|} \)?
2. How would the domain of \( f(x) = \sqrt{x} + |x| \) differ?
3. What is the range of \( f(x) = \frac{1}{\sqrt{x^2 - x - 6}} \)?
4. How do you solve quadratic inequalities like \( x^2 - x - 6 \geq 0 \)?
5. Can you explain how the absolute value function affects domains?

### Tip:
Always check for values that make the denominator zero or lead to negative values inside square roots when finding domains.

Find the domain of the following functions: (a) f(x) = 1 / (√x + |x|) (b) f(x) = 1 / √(x^2 - x - 6)

Explore how to find the domain of functions involving square roots and absolute values, such as f(x) = 1 / (√x + |x|) and f(x) = 1 / √(x^2 - x - 6). This detailed explanation covers quadratic inequalities and algebraic principles. Suitable for high school students (Grades 10-12).

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